Problems with worked out solutions about acceleration of electrons in the electric field of an Electron Gun
Problem 1)
Explain the influence of the heater voltage and the influence of the acceleration voltage on the electron beam of an Electron Gun.
Solutions:
Electrons are 'boiled off'a hot metal plate because of the Edison-Richardson-Effect and made up an 'electron gas' around the filament. The heater voltage determines the temperature of the filament. The higher the heater voltage the higher the temperature of the filament and the more electrons are 'boiled off' the filament. So more electrons can flow from the hot cathode to the anode. The electron beam is getting more intense when the heater voltage rises.
The acceleration voltage determines the electric field between hot cathode and anode. So the acceleration voltage causes the acceleration of the electrons in the direction of the anode. The higher the acceleration voltage the faster the electrons get when passing the anode.
Problem 2)
The acceleration voltage of an Electron Gun is Va = 500 V and the distance between Kathode and Anode is d = 5cm. Non-relativistic calculation can be used.
Calculate how fast the electrons are moving when they pass the anode.
Find the work done by the electric field on an electron.
What time does the electron take to get from the hot cathode to the anode?
Now the acceleration voltage changes and electrons are passing the anode with a speed of $v=2.054\cdot 10^7 \,\frac{\text m}{\text s}$. Find the acceleration voltage.
$$W_{\text{el}}=F_{\text{el}}\cdot \text {d} = V_{\text a}\cdot e \Rightarrow W_{\text{el}}=500 \,\text V \cdot 1.6\cdot 10^{-19}\,\text C \approx 8.0 \cdot 10^{-17} \,\text J$$
For uniformly accelerated motions is:
$$s=\frac{1}{2} a\cdot t^2 \Rightarrow t=\sqrt{\frac{2s}{a}}$$
The force acting on electrons in an electric field is
$$F_{\text{el}}=\frac{V_{\text a}\cdot e}{\text d}.$$ With $F=m\cdot a$ the acceleration of the elctrons is:
$$a=\frac{F_{\text{el}}}{m_{\text e}}=\frac{V_{\text a}\cdot e}{{m_{\text e}\cdot \text d}}$$ Putting this in our starting formula we find: $$t=\sqrt{\frac{2\cdot s}{\frac{V_{\text a}\cdot e}{{m_{\text e}\cdot \text d}}}}=\sqrt{\frac{2\cdot s\cdot m_{\text e}\cdot \text d}{V_{\text a}\cdot e}}\Rightarrow t=\sqrt{\frac{2\cdot 9.1\cdot 10^{-31}\,\text{kg}\cdot 0.05\,\text m\cdot 0.05\,\text m}{500\,\text V\cdot 1.6\cdot 10^{-19}\text C}} \approx 7.54\cdot 10^{-9}\,\text s$$
The acceleration voltage of an electron gun is $V_{\text a}=8.0\,\rm{kV}$.
Is non-relativistic calculation correct or must the priciples of relativity benn taken into account? Give reasons for your answer.
Calculate how fast the electrons are moving when they pass the anode.
Now the acceleration voltage changes and electrons are passing the anode with a speed of $v= \frac{1}{2}\cdot c$. Find the acceleration voltage.
Concerning the non-relativistic calculation, which acceleration voltage is needed to accelerate electrons to $v= \frac{1}{2}\cdot c$ ? What is the relative error of this calculation?
Solutions:
A relativistic calculation must be done because the acceleration voltage is higher than 2.7 kV and so the spped of the electrons is higher than 10% of the speed of light. So non-relativistic calculation lead to a non negligibly error.
non-relativistic: $$V_{\text a}=\frac{v^2\cdot m}{2\cdot e} \Rightarrow V_{\text a}=\frac{c^2\cdot m}{8\cdot e}\approx 63.984\,\text {kV}$$ relative error \(f\): $$f=\frac{V_{\rm{class}}}{V_{\rm{rel}}}-1\Rightarrow\frac{63984\,\text V}{79187\,\text V}-1\approx -0.192=-19.2\,\%$$With non-relativistic calculation the acceleration voltage would be about \(19.2\%\) too low.